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Option D) is correct

The Cyclicity table for $7$ is as follows:
$7^1 =7$
$7^2 =49$
$7^3 = 343$
$7^4 = 2401$

Hence cyclicity of $7$ is $4$

Let’s divide $78$ by $4$ and the remainder is $2$
Thus, the last digit of $7^{78}$ is equal to the last digit of $7^2$ i.e. $9$

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To get the unit digit of $7^{78}$, we will find $7^{78}\;mod\;10.$

 $7^{78}\; mod\;10=7^{76+2}\;mod\;10$

$=(7^{76}*7^{2})\;mod\;10$

$=(7^{76}\;mod\;10*7^{2}\;mod\;10)\;mod\;10$

$=((7^{4})^{19}\;mod\;10*7^{2}\;mod\;10)\;mod\;10$

$=(1*7^{2}\;mod\;10)\;mod\;10$    ( $\because$ when $a,n$ are co-primes then $a^{\Phi (n)}mod\;n=1$. Here $7,10$ are co-primes and $\Phi (10)=4$. So, $7^{4}\;mod\;10=1$ ) $[\Phi(n) =  Euler \;\;Totient\;\; Function ]$

$=(49\;mod\;10)\;mod 10$

$=9\;mod\;10$

$= 9$
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